Ideal generated by an ideal and an element
Let $R$ a commutative ring with unity. Let $anotin I$ and we consider the ideal $(I,a)$. I have already shown that $$(I,a)=big{i+ra;|;iin I, rin Rbig}.$$
Why, $(I,a)=(Icup(a))$?
I know that $(I,a)$ is the smallest ideal that contains $I$ and $a$, but how do you say that he coincides with $(Icup (a))$?
Thanks!
abstract-algebra
edited Nov 27 at 22:04
user26857
39.2k123983
asked Nov 27 at 16:57
Jack J.
4431419
add a comment |
Let $R$ a commutative ring with unity. Let $anotin I$ and we consider the ideal $(I,a)$. I have already shown that $$(I,a)=big{i+ra;|;iin I, rin Rbig}.$$
Why, $(I,a)=(Icup(a))$?
I know that $(I,a)$ is the smallest ideal that contains $I$ and $a$, but how do you say that he coincides with $(Icup (a))$?
Thanks!
abstract-algebra
edited Nov 27 at 22:04
user26857
39.2k123983
asked Nov 27 at 16:57
Jack J.
4431419
1
Why $I color{red}{cup} (a)$? Example, $R = mathbb Z$, $a = 4, I = 6mathbb Z$, then $I cup (4) = 4mathbb Z cup 6 mathbb Z$, but $J=(I, a) = 2mathbb Z$: $4, 6 in Jimplies 6-4 = 2in J$, hence $(2)subseteq J$. Since $mathbb Z$ is a PID, $J = (2)$. Then $10 in J$, but $10 notin 4mathbb Z cup 6mathbb Z$.
– xbh
Nov 27 at 17:05
I was wrong to write, $(I,a)=(Icup (a))$
– Jack J.
Nov 27 at 17:07
add a comment |
Let $R$ a commutative ring with unity. Let $anotin I$ and we consider the ideal $(I,a)$. I have already shown that $$(I,a)=big{i+ra;|;iin I, rin Rbig}.$$
Why, $(I,a)=(Icup(a))$?
I know that $(I,a)$ is the smallest ideal that contains $I$ and $a$, but how do you say that he coincides with $(Icup (a))$?
Thanks!
abstract-algebra
edited Nov 27 at 22:04
user26857
39.2k123983
asked Nov 27 at 16:57
Jack J.
4431419
Let $R$ a commutative ring with unity. Let $anotin I$ and we consider the ideal $(I,a)$. I have already shown that $$(I,a)=big{i+ra;|;iin I, rin Rbig}.$$
Why, $(I,a)=(Icup(a))$?
I know that $(I,a)$ is the smallest ideal that contains $I$ and $a$, but how do you say that he coincides with $(Icup (a))$?
Thanks!
abstract-algebra
abstract-algebra
edited Nov 27 at 22:04
user26857
39.2k123983
asked Nov 27 at 16:57
Jack J.
4431419
edited Nov 27 at 22:04
user26857
39.2k123983
asked Nov 27 at 16:57
Jack J.
4431419
edited Nov 27 at 22:04
user26857
39.2k123983
edited Nov 27 at 22:04
user26857
39.2k123983
edited Nov 27 at 22:04
user26857
39.2k123983
39.2k123983
asked Nov 27 at 16:57
Jack J.
4431419
asked Nov 27 at 16:57
Jack J.
4431419
asked Nov 27 at 16:57
Jack J.
4431419
4431419
1
Why $I color{red}{cup} (a)$? Example, $R = mathbb Z$, $a = 4, I = 6mathbb Z$, then $I cup (4) = 4mathbb Z cup 6 mathbb Z$, but $J=(I, a) = 2mathbb Z$: $4, 6 in Jimplies 6-4 = 2in J$, hence $(2)subseteq J$. Since $mathbb Z$ is a PID, $J = (2)$. Then $10 in J$, but $10 notin 4mathbb Z cup 6mathbb Z$.
– xbh
Nov 27 at 17:05
I was wrong to write, $(I,a)=(Icup (a))$
– Jack J.
Nov 27 at 17:07
add a comment |
1
Why $I color{red}{cup} (a)$? Example, $R = mathbb Z$, $a = 4, I = 6mathbb Z$, then $I cup (4) = 4mathbb Z cup 6 mathbb Z$, but $J=(I, a) = 2mathbb Z$: $4, 6 in Jimplies 6-4 = 2in J$, hence $(2)subseteq J$. Since $mathbb Z$ is a PID, $J = (2)$. Then $10 in J$, but $10 notin 4mathbb Z cup 6mathbb Z$.
– xbh
Nov 27 at 17:05
I was wrong to write, $(I,a)=(Icup (a))$
– Jack J.
Nov 27 at 17:07
1
1
Why $I color{red}{cup} (a)$? Example, $R = mathbb Z$, $a = 4, I = 6mathbb Z$, then $I cup (4) = 4mathbb Z cup 6 mathbb Z$, but $J=(I, a) = 2mathbb Z$: $4, 6 in Jimplies 6-4 = 2in J$, hence $(2)subseteq J$. Since $mathbb Z$ is a PID, $J = (2)$. Then $10 in J$, but $10 notin 4mathbb Z cup 6mathbb Z$.
– xbh
Nov 27 at 17:05
Why $I color{red}{cup} (a)$? Example, $R = mathbb Z$, $a = 4, I = 6mathbb Z$, then $I cup (4) = 4mathbb Z cup 6 mathbb Z$, but $J=(I, a) = 2mathbb Z$: $4, 6 in Jimplies 6-4 = 2in J$, hence $(2)subseteq J$. Since $mathbb Z$ is a PID, $J = (2)$. Then $10 in J$, but $10 notin 4mathbb Z cup 6mathbb Z$.
– xbh
Nov 27 at 17:05
I was wrong to write, $(I,a)=(Icup (a))$
– Jack J.
Nov 27 at 17:07
I was wrong to write, $(I,a)=(Icup (a))$
– Jack J.
Nov 27 at 17:07
add a comment |
1 Answer
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The ideal $(Icup(a))$ is the smallest ideal containing $I$ and $(a)$; in particular it contains $I$ and $a$, so it contains $I+(a)=(I,a)$. Conversely, every ideal containing $a$ also contains $(a)$, so the smallest ideal containing $I$ and $a$ also contains $I$ and $(a)$. Therefore $(I,a)$ contains $(Icup(a))$.
Hence $I+(a)=(I,a)=(Icup(a))$.
answered Nov 27 at 17:21
egreg
178k1484201
add a comment |
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The ideal $(Icup(a))$ is the smallest ideal containing $I$ and $(a)$; in particular it contains $I$ and $a$, so it contains $I+(a)=(I,a)$. Conversely, every ideal containing $a$ also contains $(a)$, so the smallest ideal containing $I$ and $a$ also contains $I$ and $(a)$. Therefore $(I,a)$ contains $(Icup(a))$.
Hence $I+(a)=(I,a)=(Icup(a))$.
answered Nov 27 at 17:21
egreg
178k1484201
add a comment |
The ideal $(Icup(a))$ is the smallest ideal containing $I$ and $(a)$; in particular it contains $I$ and $a$, so it contains $I+(a)=(I,a)$. Conversely, every ideal containing $a$ also contains $(a)$, so the smallest ideal containing $I$ and $a$ also contains $I$ and $(a)$. Therefore $(I,a)$ contains $(Icup(a))$.
Hence $I+(a)=(I,a)=(Icup(a))$.
answered Nov 27 at 17:21
egreg
178k1484201
add a comment |
The ideal $(Icup(a))$ is the smallest ideal containing $I$ and $(a)$; in particular it contains $I$ and $a$, so it contains $I+(a)=(I,a)$. Conversely, every ideal containing $a$ also contains $(a)$, so the smallest ideal containing $I$ and $a$ also contains $I$ and $(a)$. Therefore $(I,a)$ contains $(Icup(a))$.
Hence $I+(a)=(I,a)=(Icup(a))$.
answered Nov 27 at 17:21
egreg
178k1484201
The ideal $(Icup(a))$ is the smallest ideal containing $I$ and $(a)$; in particular it contains $I$ and $a$, so it contains $I+(a)=(I,a)$. Conversely, every ideal containing $a$ also contains $(a)$, so the smallest ideal containing $I$ and $a$ also contains $I$ and $(a)$. Therefore $(I,a)$ contains $(Icup(a))$.
Hence $I+(a)=(I,a)=(Icup(a))$.
answered Nov 27 at 17:21
egreg
178k1484201
answered Nov 27 at 17:21
egreg
178k1484201
answered Nov 27 at 17:21
egreg
178k1484201
answered Nov 27 at 17:21
egreg
178k1484201
178k1484201
add a comment |
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Why $I color{red}{cup} (a)$? Example, $R = mathbb Z$, $a = 4, I = 6mathbb Z$, then $I cup (4) = 4mathbb Z cup 6 mathbb Z$, but $J=(I, a) = 2mathbb Z$: $4, 6 in Jimplies 6-4 = 2in J$, hence $(2)subseteq J$. Since $mathbb Z$ is a PID, $J = (2)$. Then $10 in J$, but $10 notin 4mathbb Z cup 6mathbb Z$.
– xbh
Nov 27 at 17:05
I was wrong to write, $(I,a)=(Icup (a))$
– Jack J.
Nov 27 at 17:07